[/quote]
,
.
and 
Prove that
.
such that for all 
,![\[
f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]](http://latex.artofproblemsolving.com/f/9/3/f93b7526ff4338eea89a1c59bc009aa001ffc15d.png)
Prove that
be an acute triangle with 
and orthocenter 
.
and 
be the feet of the altitudes from 
and 
onto sides 
and 
,
be the midpoint of 
,
be the midpoint of 
.
meet 
and 
.
meets 
and 
.
be the second intersection point (different from 
and 
.
and 
Prove that
as its incenter and 
as its circumcircle, 
intersects 
.
be a point on the arc 
,
a point on the segment 
.
is the midpoint of 
,
and 
lies on 
and 
be the incircle and circumcircle of the acute triangle 
so that all of its sides are tangent to 
,
are both on 
and 
,
,
and 
intersects on 
as its circumcentre, 
as its nine-point centre.
,
and 
to be the midpoint of the smaller arcs.
as the ninja point of the contact triangle
Points 
Points 
-
be the reflection of 
is tangent to the incircle.
be an odd prime number. Suppose 
,![\[
1 + \cfrac{x}{1 + \cfrac{2x}{1 + \cfrac{\ddots}{1 +
(p-1)x}}}=\frac{P(x)}{Q(x)}.
\]](http://latex.artofproblemsolving.com/b/c/a/bca9034e2affa1d80b167451d77434fbcd8bd88a.png)
Show that all coefficients of 
,
and 
,
a right-angled triangle can be formed?
Y by kiyoras_2001, Adventure10, GeoKing, Mango247, and 1 other user
Given a triangle 
. The angle bisectors of the angles and 
intersect the sides 
and 
at the points 
and 
, and intersect each other at the point 
. The line 
intersects the circumcircle of triangle at the points 
and 
. Prove that the circumradius of triangle 
is twice as long as the circumradius of triangle .
. The angle bisectors of the angles and 
intersect the sides 
and 
at the points 
and 
, and intersect each other at the point 
. The line 
intersects the circumcircle of triangle at the points 
and 
. Prove that the circumradius of triangle 
is twice as long as the circumradius of triangle .This post has been edited 2 times. Last edited by darij grinberg, Oct 24, 2007, 8:19 PM
and 
and for the points 
define the points 
Then for any point 
there is the following relation : ![\[ \boxed {\left|b\cdot \frac{YB}{YA}\cdot \frac{1}{DB}-c\cdot \frac{XC}{XA}\cdot \frac{1}{DC}\right|=a\cdot \frac{1}{AD}}\ . \]](http://latex.artofproblemsolving.com/5/3/f/53ffc50d333747a6f996b00852514f3469298dad.png)
Remark. If 
then we obtain the above problem from Bulgaria, 1997.
. I have a pretty short and simple proof, but it took me a while to see what was going on, and I used dynamic geometry software.
denote the 
passes through these two points we're done, because then 
,
is the orthic axis of 
,
,
.
and 
are two given distinct circles. A point 
such that the tangent at 
at two points 
of 
and 
,
is a constant in term of 
.
.
.
respectively, then the circumcenter 
which lie on a circle center 
at the two circle's intersections. Then without difficulty, you can see that the midpoints of 
and 
are the circumcenter 
with ratio 
maps these midpoints to 
.
.
to 
to 
as 
and 
are cyclic. Call them 
and 
.
is the radical center of 
and 
is the radical center of 
so 
are on the radical axis of 
are the intersections of 
is cyclic. Now because of the homothety taking 
be the midpoints of minor arcs 
be the 
excenters. By PoP, we have that 
so 
,
.
and 
,
and 
are congruent so the circumradii are equal too, so 
respective excenters of 
Since 
is cyclic, point 
therefore 
We are done, because in triangle 
is the orthocenter;
is the nine-point circle.
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as the 
excenters
as 
as midpoint of arc 
respectively.
is twice the circumradius of 
is the midpoint of 
and 
is the midpoint of 
.
.
the circumradius of 
are concurrent
Note that 
But note that 
.
.
is cyclic. So by POP, we have 
And we are done as The circumradius of 
is twice the circumradius of 
and 
be the 
,
,
both lie on 
,
denote the 
analogously. Let line 
meet 
at point 
where the last equality follows due to the existence of 
.
,
.
.
it follows that 
have equal radius. Finally, remark that 
is the nine point circle of 
